Chapter 4

Question 1

Why do we use numerical integration in modelling?

Answer 1

Numerical integration is the approximation of a continuous model in discrete steps (both in time and space for spatial models)

Question 2

What are the advantages of using numerical integration for modelling instead of analytical solutions?

Answer 2

1. Many models are not solvable analytically, but always numerically
2. With numerical integration, there is virtually no limit to the complexity of models that can be solved:
   * Driving variables (e.g. time series of weather or deposition)
   * Discrete events (e.g. harvesting)

Question 3

Equation for logistic growth rate?

Answer 3

dN/dt = r * N * (1 - N/K)
K = carrying capacity

Question 4

What is a driving variable?

Answer 4

Driving variables characterize the influence of external factors on the system and may thus be essential to the model, yet they are not influenced by the processes within the system and thus not explicitly modelled.

Question 5

What do Ordinary Differential Equations describe?

Answer 5

The change in states

Question 6

How can be ODEs mathematically represented?

Answer 6

Y’ = f(Y, p, t)

Question 7

What is a condition to solve this equation Y’ = f(Y, p, t)?

Answer 7

The initial state needs to be known (state Y at time = 0, Y_0)

Question 8

What is the mathematical representation of the forward finite difference method?

Answer 8

Y_t+dt = Y_t + dt * f(Y, p, t)

Question 9

In a logistic growth model describe what happens when N (state) is very small

Answer 9

dN/dt = r * N * (1 - N/K)
if N= very small (0) then eq becomes
dN/dt = r * N * 1
meaning that the growth rate is now exponential

Question 10

In a logistic growth model describe what happens when N (state) is equal or approaching K

Answer 10

dN/dt = r * N * (1 - N/K)
if N= K then eq becomes
dN/dt = r * N * (1 - K/K)
= r * N * (1 - 1) =
= r * N * 0 =
= 0
meaning that the growth rate is now 0 thus the population stops growing

Question 11

In a logistic growth model describe what happens when N (state) is bigger than K

Answer 11

dN/dt = r * N * (1 - N/K)
if N > K then eq becomes
dN/dt = r * N * (1 - 2)
= r * N * - (value) –> whole eq become negative, thus the growth rate will also be negative

Question 12

What is the analytical solution to a simple logistic model?

Answer 12

N_t = (K * N_0 * e ^ (r *t)) / (K + N_0 * (e ^ (r * t) - 1))

Question 13

What are the components used in the ode() function in R?

Answer 13

1. the value(s) of initial state variable(s) – Y0
2. the value(s) of parameter(s) – p
3. a function with the auxiliary and differential equations returning the rates of all state variables –
   f
4. time – t

Question 14

When is the highest growth in a logistical equation?

Answer 14

when N = K/2

Question 15

What happens when the time step we chose for the model is too large?

Answer 15

The model oscillates around the equilibrium

Question 16

What is the order of variables when inputting in function ode()?

Answer 16

ode(y = state
t = time sequence
func = logistic growth function
parms = parameters
method = euler’s/runge-kutta)

Question 17

What should event data frames contain?

Answer 17

• var: the state variable name (or number) that is affected by the event;
• time: the time at which the event is to take place;
• value: the value, magnitude of the event;
• method: which event is to take place; it should be one of “replace”, “add”, “multiply” (but it is also allowed to specify the number 1 = replace, 2 = add, 3 = multiply)

Question 18

How are driving variables included in the ode() function?

Answer 18

They are included through forcing functions.

Question 19

What are forcing functions?

Answer 19

Functions that appear in the equations and are only a function of time, and not of any of the other variables.

Question 20

When dealing with driving variables in a model, how do we usually have data and what should we do to make it continuous?

Answer 20

Often, we have data on driving variables gathered as a time series, containing values only at specified times.

Thus, if a model uses driving variables, their value at each time point needs to be estimated by interpolation of the data series.

Question 21

How does the 4th-order Runge-Kutta differ from Euler’s method of integration?

Answer 21

4th-order Runge-Kutta is a weighted form of Euler’s with different rates of change. RK4 involves four function evaluations per step. The different rates of change are calculated from the initial starting point. First, the rate of change is calculated for half a time step and the final and fourth one is calculated for the whole time step. Makes the approximation better.
…

Question 22

How do we measure the approximation of the model?

Answer 22

1. Accuracy (small numerical error)
2. Efficiency (fast to solve)
3. Stability (no oscillations or chaotic behaviour)
4. Memory usage

Question 23

How to make the approximation better?

Answer 23

1. Make Δt smaller
2. Evaluation at multiple points in time during ∆t
3. Including higher order derivatives e.g. using a Taylor expansion
4. Using a variable time-step instead of a fixed time-step
5. Small time step when dynamics are fast
6. Large time step when dynamics are slow

Question 24

Euler’s and Runge-Kutta mathematical equations

Answer 24

1. Euler’s
   y_t+dt = y_t + dt * dy/dt
2. Runge-kutta
   y_t + dt = y_t + 1/6 * dt * (k_1 + 2k_2 + 2k_3 + k_4)